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Find Tangent Slope with Polar coordinates

  1. May 9, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the slope of the tangent line to the given polar curve at the point specified by the value of θ.
    r = 9sin(θ)
    θ = pi/6

    2. Relevant equations
    dy/dx = (dy/dθ) / (dx/dθ)
    x=rcosθ
    y=rsinθ

    (sinx)^2 = (1/2)(1-cos2x)
    (cosx)^2 = (1/2)(1+cos2x)
    2sinxcosx = sin(2x)

    3. The attempt at a solution
    r = 9sin(θ)
    x=rcos(θ)=9sin(θ)cos(θ)= --> Would this be sin(9θ) or sin(18θ)?
    y=rsin(θ)=9sin(θ)sin(θ)= 9sin^2(θ)


    dy/dx = (dy/dθ) / (dx/dθ) = (9 * 9sinθcosθ) / (cos(18θ) * 9) = sin(18θ)/cos(18θ) = tan(18θ)

    θ = pi/6
    tan(18*pi/6)=tan(3pi)=0
     
  2. jcsd
  3. May 9, 2010 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hi JSGhost! Welcome to PF! :smile:

    (have a pi: π and try using the X2 tag just above the Reply box :wink:)
    Nooo! you can't move numbers in and out of a function like sin (you wouldn't do it for √, would you? :wink:)
    But you're not differentiating. :confused:

    Start again. :smile:
     
  4. May 9, 2010 #3
    Thanks for replying. I actually figured it out after posting this topic. Saw a similar problem to this one on another site. Thanks anyways.

    Find the slope of the tangent line to the given polar curve at the point specified by the value of θ.
    r = 9sin(θ)
    θ = pi/6

    x=rcos(θ)=9sin(θ)cos(θ)= 9(sin(2θ))
    y=rsin(θ)=9sin(θ)sin(θ)= 9sin^2(θ)

    dx/dθ = 9(cos^2(θ) - sin^2(θ))
    dy/dθ = 9(2sinθcosθ)

    dy/dx = (dy/dθ)/(dx/dθ)
    = 9(2sinθcosθ) / 9(cos^2(θ) - sin^2(θ))
    = [2sinθcosθ] / [cos^2(θ)-sin^2(θ)]

    when θ = pi/6
    dy/dx = [2sin(pi/6)cos(pi/6)] / [cos^2(pi/6) - sin^2(pi/6)]
    = [2*(1/2)*(sqrt(3)/2)] / [(sqrt(3)/2)*(sqrt(3)/2)-(1/2)*(1/2)]
    = (sqrt(3)/2) / 2/4 = (sqrt(3)/2) / (1/2) = sqrt(3)
     
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